What is the integral of (2x^3-8x+5) ?

The integral of (2x^3-8x+5) is (x^4)/2-4x^2+5x+C

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integral of (2x^3-8x+5)

\int (2 x^{3} - 8 x + 5) d x

\frac{x ^{4}}{2} - 4 x^{2} + 5 x + C

Solution steps

\int 2 x^{3} - 8 x + 5 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 2 x^{3} d x - \int 8 x d x + \int 5 d x

\int 2 x^{3} d x = \frac{x ^{4}}{2}

\int 8 x d x = 4 x^{2}

\int 5 d x = 5 x

= \frac{x ^{4}}{2} - 4 x^{2} + 5 x

Add a constant to the solution

= \frac{x ^{4}}{2} - 4 x^{2} + 5 x + C

\int + \frac{x ^{2} - 81}{x ^{3}} d x

\int \frac{2 x ^{3} + 5 x}{3 x ^{5}} d x

\int \frac{x ^{3}}{x ^{2} + 8} d x

\int \frac{1}{y ( 1 + ln ( y ) )} d y

\int + 4 x e^{- x} d x

What is the integral of (2x^3-8x+5) ?
The integral of (2x^3-8x+5) is (x^4)/2-4x^2+5x+C